Question

4c₃h₅n₃o₉ → 12co₂ + 6n₂ + o₂ + 10h₂o
2.50 kg of nitroglycerin detonates. how many kg of co₂ and h₂o are produced?

Explanation:

Step1: Convert mass of nitroglycerin to moles

First, find the molar mass of $C_3H_5N_3O_9$.
$M_{C_3H_5N_3O_9}=3\times12.01 + 5\times1.01+3\times14.01 + 9\times16.00=227.09\ g/mol$.
The mass of nitroglycerin is $m = 2.50\ kg=2500\ g$.
The number of moles of nitroglycerin, $n=\frac{m}{M}=\frac{2500\ g}{227.09\ g/mol}\approx11.01\ mol$.

Step2: Use mole - ratio to find moles of $CO_2$

From the balanced equation $4C_3H_5N_3O_9
ightarrow12CO_2 + 6N_2+O_2 + 10H_2O$, the mole - ratio of $C_3H_5N_3O_9$ to $CO_2$ is $\frac{n_{CO_2}}{n_{C_3H_5N_3O_9}}=\frac{12}{4} = 3$.
So, $n_{CO_2}=3\times n_{C_3H_5N_3O_9}=3\times11.01\ mol = 33.03\ mol$.
The molar mass of $CO_2$ is $M_{CO_2}=12.01+2\times16.00 = 44.01\ g/mol$.
The mass of $CO_2$, $m_{CO_2}=n_{CO_2}\times M_{CO_2}=33.03\ mol\times44.01\ g/mol\approx1453\ g = 1.45\ kg$.

Step3: Use mole - ratio to find moles of $H_2O$

The mole - ratio of $C_3H_5N_3O_9$ to $H_2O$ is $\frac{n_{H_2O}}{n_{C_3H_5N_3O_9}}=\frac{10}{4}=2.5$.
So, $n_{H_2O}=2.5\times n_{C_3H_5N_3O_9}=2.5\times11.01\ mol = 27.525\ mol$.
The molar mass of $H_2O$ is $M_{H_2O}=2\times1.01+16.00 = 18.02\ g/mol$.
The mass of $H_2O$, $m_{H_2O}=n_{H_2O}\times M_{H_2O}=27.525\ mol\times18.02\ g/mol\approx496\ g = 0.496\ kg$.

Answer:

The mass of $CO_2$ produced is $1.45\ kg$ and the mass of $H_2O$ produced is $0.496\ kg$.